Question

The value of tan A +sin A=M and tan A – sin A=N .

The value of $\frac{(M^2-N^2)}{(MN{)}^{0.5}}$

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is D

4

$M^2$-$N^2$ = (Tan A+ Sin A + Tan A –Sin A)(Tan A +Sin A – Tan A+ Sin A)

$M^2$-$N^2$ =4 tan A sin A

and $(MN{)}^{0.5}$ = $(ta{n}^{2}A-si{n}^{2}A{)}^{0.5}$

$(MN{)}^{0.5}$ =$sinA\left[\right(\frac{1}{co{s}^{2}A})-1{]}^{0.5}$

$(MN{)}^{0.5}$ =tan A sin A

Therefore, $\frac{(m^2-n^2)}{(mn{)}^{0.5}}$= 4